
The characteristics of Hydrogen resemble the elements of _________ as well as ________ groups
A.IIA and VA
B.IIIA and VIA
C.IIA and VIA
D.1A and VIIA
Answer
578.7k+ views
Hint: Write down the general outermost electronic configuration of the elements in these groups and also write the outermost electronic configuration of hydrogen atom. Then compare Hydrogen configuration with the other groups on the basis of outermost shell configuration. Note the number of electrons which need to be gained or lost in order to achieve octet and compare it with Hydrogen.
Complete step by step answer:
Hydrogen has the outermost shell configuration as $1{s^1}$ and hence it can attain octet configuration either by losing one electron and becoming $1{s^0}$ and forming a proton $[{H^ + }]$ or by gaining an electron and having the outermost shell configuration as $1{s^2}$ and forming a Hydride Ion $[{H^ - }]$.
In both these cases, Hydrogen attains stability by a $ \pm 1$ change in the outermost electron count.
$\therefore $ Hydrogen will have properties similar to that of those group elements which will also attain stability either by losing one electron or by gaining one electron. Since there is no other element like hydrogen in the periodic table which can attain stability by either gaining or losing an electron. Hence Hydrogen Shows similarity with 2 separate groups rather than 1 individual group.
The group of elements which can attain stability by losing one electron belong to the first group which is IA. This group consists of Alkali Earth metals such as $N{a^ + },{K^ + },C{s^ + }$ with their outermost electronic configuration as $n{s^1}$
Whereas Group VIIA elements can attain octet by gaining one electron. This group is Known as the Halogen family and consists of elements such as ${F^ -},C{l^ - },B{r^ - }$ and their outermost electronic configuration is $n{s^2}n{p^5}$.
Hence option D is correct.
Note:
Only compare the ground state electronic configuration of Hydrogen with the General ground state electronic configuration. Do not compare with the Excited state electronic configuration and do not use Ionisation Enthalpy and Electron Gain Enthalpy Values as a method of comparison since due to the size of the atoms, these values would differ greatly.
Complete step by step answer:
Hydrogen has the outermost shell configuration as $1{s^1}$ and hence it can attain octet configuration either by losing one electron and becoming $1{s^0}$ and forming a proton $[{H^ + }]$ or by gaining an electron and having the outermost shell configuration as $1{s^2}$ and forming a Hydride Ion $[{H^ - }]$.
In both these cases, Hydrogen attains stability by a $ \pm 1$ change in the outermost electron count.
$\therefore $ Hydrogen will have properties similar to that of those group elements which will also attain stability either by losing one electron or by gaining one electron. Since there is no other element like hydrogen in the periodic table which can attain stability by either gaining or losing an electron. Hence Hydrogen Shows similarity with 2 separate groups rather than 1 individual group.
The group of elements which can attain stability by losing one electron belong to the first group which is IA. This group consists of Alkali Earth metals such as $N{a^ + },{K^ + },C{s^ + }$ with their outermost electronic configuration as $n{s^1}$
Whereas Group VIIA elements can attain octet by gaining one electron. This group is Known as the Halogen family and consists of elements such as ${F^ -},C{l^ - },B{r^ - }$ and their outermost electronic configuration is $n{s^2}n{p^5}$.
Hence option D is correct.
Note:
Only compare the ground state electronic configuration of Hydrogen with the General ground state electronic configuration. Do not compare with the Excited state electronic configuration and do not use Ionisation Enthalpy and Electron Gain Enthalpy Values as a method of comparison since due to the size of the atoms, these values would differ greatly.
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